R50 · Thin lens equation: image distance and magnification

This problem uses the thin lens equation to determine the image characteristics.

### Given Values:
*   Focal length ($f$) = $+10 \text{ cm}$ (Positive because it is a converging lens)
*   Object distance ($d_o$) = $15 \text{ cm}$

---

## (a) Calculate the Image Distance ($d_i$)

We use the thin lens equation:
$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Rearrange the equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the known values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract the fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30 \text{ cm}}\right) - \left(\frac{2}{30 \text{ cm}}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30 \text{ cm}}$$
$$\frac{1}{d_i} = \frac{1}{30 \text{ cm}}$$

Invert the result to find $d_i$:
$$d_i = +30 \text{ cm}$$

**The image distance ($d_i$) is $+30 \text{ cm}$.**

---

## (b) Calculate the Magnification ($m$)

We use the magnification equation:
$$m = -\frac{d_i}{d_o}$$

Substitute the values for $d_i$ and $d_o$:
$$m = -\frac{(+30 \text{ cm})}{(15 \text{ cm})}$$
$$m = -2$$

**The magnification ($m$) is $-2$.**

---

## Conclusion and Image Characteristics

| Characteristic | Value | Interpretation |
| :--- | :--- | :--- |
| **Image Distance ($d_i$)** | $+30 \text{ cm}$ | Positive sign means the image forms on the opposite side of the lens from the object. The image is **Real**. |
| **Magnification ($m$)** | $-2$ | Negative sign indicates that the image is flipped relative to the object. The image is **Inverted**. |

The resulting image is **REAL** and **INVERTED**. (It is also magnified, as $|m|=2$).

To solve the problem, we will apply the

To solve this problem, we will use the given values and the thin lens equation.

### **Given Data:**
*   Focal length ($f$) = $+10\text{ cm}$ (positive because it is a converging lens)
*   Object distance ($d_o$) = $15\text{ cm}$

---

### **(a) Calculate the image distance ($d_i$)**

Using the thin lens equation:
$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Rearrange the equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10} - \frac{1}{15}$$

To subtract these fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \frac{3}{30} - \frac{2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30}$$

Invert both sides to find $d_i$:
**$d_i = 30\text{ cm}$**

---

### **(b) Calculate the magnification ($m$)**

Using the magnification formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the values:
$$m = -\frac{30\text{ cm}}{15\text{ cm}}$$
**$m = -2$**

---

### **Final Image Characteristics**

Based on the calculations above, we can determine the nature of the image:

1.  **Real or Virtual?** Since the image distance ($d_i$) is **positive** ($+30\text{ cm}$), the image is **real**. (Real images are formed on the opposite side of the lens from the object).
2.  **Upright or Inverted?** Since the magnification ($m$) is **negative** ($-2$), the image is **inverted**.
3.  **Size:** Since the absolute value of the magnification is greater than 1 ($|m| > 1$), the image is **enlarged** (twice the size of the object).

**Summary:** The image is **real, inverted, and enlarged**.

This problem uses the thin lens equation to determine the location and characteristics of an image formed by a convex lens.

### Given Values:
*   Focal length ($f$): $+10 \text{ cm}$ (Positive for a converging lens)
*   Object distance ($d_o$): $+15 \text{ cm}$
*   Thin Lens Equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

***

### (a) Calculate the image distance $d_i$

We rearrange the thin lens equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract the fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30}\right) - \left(\frac{2}{30}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30 \text{ cm}}$$

Invert the expression to find $d_i$:
$$d_i = +30 \text{ cm}$$

**The image distance ($d_i$) is $+30 \text{ cm}$.**

***

### (b) Calculate the magnification $m$

We use the magnification formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the calculated and given values:
$$m = -\frac{(+30 \text{ cm})}{(+15 \text{ cm})}$$
$$m = -2$$

**The magnification ($m$) is $-2$.**

***

### Conclusion and Image Characteristics

| Parameter | Value | Interpretation |
| :--- | :--- | :--- |
| **Image Distance ($d_i$)** | $+30 \text{ cm}$ | Since $d_i$ is positive, the image forms on the opposite side of the lens from the object. $\implies$ **Real** |
| **Magnification ($m$)** | $-2$ | The negative sign indicates the image is inverted. $|m|=2$ means it is twice as large. $\implies$ **Inverted** |

The resulting image is a **real**, **inverted**, and magnified image, located $30 \text{ cm}$ from the lens on the side opposite the object.

## Step 1: Given information
The focal length of the lens (f) is given as 10 cm. The object distance (d_o) is given as 15 cm.

## Step 2: Using the thin lens equation to find d_i
We use the formula 1/f = 1/d_o + 1/d_i, where f = 10 cm and d_o = 15 cm.

## Step 3: Substitute values into the thin lens equation
Substituting the given values into the thin lens equation gives us 1/10 = 1/15 + 1/d_i.

## Step 4: Solve for d_i
To solve for d_i, we first need to get a common denominator on the right side of the equation. This results in (d_o * d_i) / (d_o * d_i) being equal to 1/10 = 1/15 + 1/d_i.

## Step 5: Multiply both sides by d_o * d_i
After multiplying, we get d_i / (d_o * d_i) = 3*d_i / (15*d_i) + (d_o*d_i) / (d_o*d_i), which simplifies to 1/10 = 1/15 + 1/d_i.

## Step 6: Find a common denominator and simplify
Finding a common denominator of 150 for both fractions on the right side, we get 15/d_i = 10/d_i + 10/150.

## Step 7: Combine like terms
Combining the terms gives us 15/d_i = 10/d_i + 1/15*d_i. Multiplying everything by d_i to clear the fraction, we get 15 = 10 + (d_i)/15.

## Step 8: Solve for d_i
Subtracting 10 from both sides, we have 5 = (d_i)/15.

## Step 9: Multiply both sides by 15
Multiplying both sides of the equation by 15 gives us d_i = 5*15.

## Step 10: Calculate d_i value
The calculation results in a positive value for d_i, indicating that the image is real. The calculated d_i is 75 cm.

## Step 11: Determine magnification and orientation
Since d_i is positive, the image is real. The magnification m = -d_i/d_o, so we calculate it as follows: m = -(75) / (15).

## Step 12: Calculate magnification value
The calculation results in a negative value for magnification, indicating that the image is inverted.

## Step 13: State whether image is real or virtual and upright or inverted
Based on the calculations, we conclude that the resulting image is real and inverted.


The final answer is: $\boxed{75}$

Given:
- Focal length \( f = 10 \) cm
- Object distance \( d_o = 15 \) cm

We need to find the image distance \( d_i \) using the thin lens equation:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

Rearranging for \( d_i \):

\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]

Substitute the given values:

\[ \frac{1}{d_i} = \frac{1}{10 \, \text{cm}} - \frac{1}{15 \, \text{cm}} \]

Calculate each term:

\[ \frac{1}{10 \, \text{cm}} = 0.1 \]
\[ \frac{1}{15 \, \text{cm}} = \frac{1}{15} \approx 0.0667 \]

So,

\[ \frac{1}{d_i} = 0.1 - 0.0667 = 0.0333 \]

Taking the reciprocal to find \( d_i \):

\[ d_i = \frac{1}{0.0333} \approx 30 \, \text{cm} \]

So, \( d_i \) is approximately 30 cm.

(b) To calculate the magnification \( m \):

\[ m = -\frac{d_i}{d_o} \]

Substitute the values:

\[ m = -\frac{30 \, \text{cm}}{15 \, \text{cm}} = -2 \]

The magnification is -2, which means the image is inverted and twice as large as the object.

### Image Characteristics:
- The image distance \( d_i \) is positive, indicating a real image.
- The magnification \( m \) is negative, indicating that the image is inverted.

Therefore, the resulting image is real and inverted.

Given a converging lens with a focal length of 10 cm, an object is placed 15 cm in front of the lens on the optical axis. We need to find the image distance using the thin lens equation and then calculate the magnification.

First, we use the thin lens equation:
\[
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
\]
where \( f = 10 \) cm and \( d_o = 15 \) cm. Plugging in these values:

\[
\frac{1}{10} = \frac{1}{15} + \frac{1}{d_i}
\]

Subtracting \(\frac{1}{15}\) from both sides:

\[
\frac{1}{d_i} = \frac{1}{10} - \frac{1}{15}
\]

Converting to a common denominator (30):

\[
\frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30}
\]

So,

\[
\frac{1}{d_i} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}
\]

Thus, the image distance \( d_i \) is:

\[
d_i = 30 \text{ cm}
\]

Next, we calculate the magnification \( m \):

\[
m = -\frac{d_i}{d_o} = -\frac{30}{15} = -2
\]

The negative magnification indicates that the image is inverted. Since the image distance is positive, the image is real.

### Final Answer
(a) The image distance is \boxed{30} cm.  
(b) The magnification is \boxed{-2}, indicating a real and inverted image.

This problem uses the thin lens equation to determine where an image forms and what its characteristics are.

### Given Values:
*   Focal length ($f$): $+10 \text{ cm}$ (Positive because it is a converging lens)
*   Object distance ($d_o$): $15 \text{ cm}$
*   Thin Lens Equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

---

### (a) Calculate the image distance $d_i$

We rearrange the thin lens equation to solve for $1/d_i$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract these fractions, we find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30}\right) - \left(\frac{2}{30}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30}$$

$$d_i = +30 \text{ cm}$$

**The image distance $d_i$ is $+30 \text{ cm}$.**

---

### (b) Calculate the magnification $m$

We use the formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the calculated $d_i$ and the given $d_o$:
$$m = - \frac{30 \text{ cm}}{15 \text{ cm}}$$
$$m = -2$$

**The magnification $m$ is $-2$.**

---

### Image Classification

Based on the signs of $d_i$ and $m$:

1.  **Real or Virtual?**
    *   Since $d_i$ is positive ($+30 \text{ cm}$), the image forms on the same side as the object (the opposite side of the lens). Therefore, the image is **real**.

2.  **Upright or Inverted?**
    *   The magnification $m$ is negative ($-2$). A negative magnification indicates that the image is flipped relative to the object. Therefore, the image is **inverted**.

*(Note: The magnitude $|m|=2$ also tells us that the image is twice the size of the original object.)*

Given that the lens has a focal length ($f$) of $10 \text{ cm}$ and the object distance ($d_{\mathrm O}$) is $15 \text {cm}$, we can use the thin lens equation to find image distance ($d_{\mathrm I}$):

$$

\begin{equation*}
    \frac{1}{f} = \frac{1}{d_{\mathrm{O}}} + \frac{1}{d_{\mathrm{I}}}\, .
\end{equation*}
$$

Plugging in the values:

$$

\begin{align*}
    \frac{1}{0.10} &= \frac{1}{0.15} + \frac{1}{d_{\mathrm I}}\\
    10 &= 6.67 + \frac{1}{d_{\mathrm I}} \\
    3.33 & = \frac{1}{d_{\mathrm I}} \\
    d_{\mathrm I} & = 0.30 \text {cm}\, .\\
\end{align*}
$$

(a) The image distance ($d_{\mathrm I}$) is $0.30$ cm behind the lens. 

Now we can calculate magnification ($m$) using:

$$

\begin{equation*}
    m = - \frac{d_{\mathrm{I}}}{d_{\mathrm O}}\, .
\end{equation*}
$$

Plugging in the values: 

$$

\begin{align*}
   m & = -\frac{0.30 \text { cm}}{15 \text {cm}}\\
     &= -0.02 \\
     &= -1/50 \, .
\end{align*}
$$

(b) The magnification ($m$) is $-1/50$, which means the resulting image is real and inverted (since negative sign indicates inversion).

<think>
Okay, so I need to solve this problem about a converging lens. Let me start by recalling what I know. The question is about a convex lens with a focal length of 10 cm. The object is placed 15 cm in front of the lens, and they want me to find the image distance and the magnification. Also, determine if the image is real or virtual and upright or inverted. Alright, let's start with the thin lens equation. 

The thin lens equation is 1/f = 1/d_o + 1/d_i. Here, f is the focal length, d_o is the object distance, and d_i is the image distance. The magnification is given by m = -d_i/d_o. 

So, given values are f = 10 cm and d_o = 15 cm. Let me write that down:

f = 10 cm
d_o = 15 cm

First, part (a) asks for the image distance d_i. I need to rearrange the thin lens equation to solve for d_i. The equation is:

1/f = 1/d_o + 1/d_i

So, rearranging to solve for 1/d_i gives:

1/d_i = 1/f - 1/d_o

Then, taking reciprocals, d_i = 1 / (1/f - 1/d_o). Let me plug in the numbers.

First, let me compute 1/f and 1/d_o:

1/f = 1/10 cm⁻¹

1/d_o = 1/15 cm⁻¹

Then, 1/d_i = (1/10) - (1/15). To subtract these fractions, I need a common denominator. The least common multiple of 10 and 15 is 30. So:

1/10 = 3/30

1/15 = 2/30

So, 3/30 - 2/30 = 1/30. Therefore, 1/d_i = 1/30 cm⁻¹, so d_i is 30 cm. Wait, that seems straightforward. Let me check my steps again. 

Alternatively, maybe I should use another approach. Let me write it step by step:

Starting with 1/f = 1/d_o + 1/d_i,

Plugging in the numbers:

1/10 = 1/15 + 1/d_i

So, subtract 1/15 from both sides:

1/d_i = 1/10 - 1/15

Find a common denominator for 10 and 15, which is 30.

1/10 = 3/30 and 1/15 = 2/30,

Therefore, 1/d_i = 3/30 - 2/30 = 1/30,

Thus, d_i = 30 cm. 

Okay, that seems correct. So the image is formed 30 cm on the opposite side of the lens from the object. Since the object is placed in front of the lens (d_o is positive), and the image distance is also positive, which for a converging lens, a positive image distance means a real image. So that's part (a).

Now, moving on to part (b), the magnification m = -d_i/d_o.

Plugging in the values we have:

m = - (30 cm) / (15 cm) = -2. 

So the magnification is -2. The negative sign indicates that the image is inverted relative to the object. The magnification's absolute value is 2, so the image is twice as tall as the object.

Now, putting this all together, the image is real because the image distance is positive (for a converging lens, real images are on the opposite side of the lens from the object). Since the magnification is negative, the image is inverted. 

Wait, let me just verify the rules again to make sure. For converging lenses:

- When the object is beyond the focal point (so beyond 2F? Wait, actually, the focal length is 10 cm here. Wait, the object is at 15 cm, which is beyond the focal length (which is 10 cm). Hmm, actually, for a converging lens, when the object is beyond the focal length but not at infinity, the image is real and inverted. 

Wait, actually, the critical point is the object distance versus the focal length. Let me recall the different cases:

For a converging lens:

- If the object is outside the focal length (d_o > f), the image is real, inverted, and on the opposite side.

- If the object is inside the focal length (d_o < f), the image is virtual, upright, and on the same side as the object.

In this case, the object is at 15 cm, which is greater than the focal length of 10 cm, so the image is real and inverted. Which matches with our calculations (d_i = 30 cm, positive, real image, and magnification negative, inverted). 

So the magnification being -2 says inverted and twice the size. 

Therefore, the answers are (a) d_i = 30 cm, real image, (b) magnification is -2, inverted.

Wait, the problem statement says "state whether the resulting image is real or virtual, and upright or inverted." So they want that in the answers for each part? The question is part a and b are separate, but they might want the characteristics mentioned in each part? Let me check the original question again:

"calculate: (a) the image distance d_i, and (b) the magnification m = -d_i/d_o. State whether the resulting image is real or virtual, and upright or inverted."

Hmm, the way it's written, the "state" part might be for both parts, but perhaps they just want the statements along with the answers. Since in the problem statement, when they say "calculate (a)... (b)... and state...". Probably, they just want the characteristics mentioned in the answer. 

So in summary, for part a, image distance is 30 cm, real. For part b, magnification is -2, which means inverted and magnified by 2 times. The image is real and inverted. 

Wait, just to ensure that the image is indeed inverted. Since the magnification is negative, that's the standard convention, right? The magnification's sign indicates orientation. So if m is negative, image is inverted. 

Alternatively, sometimes people might get confused, but in the formula m = -d_i/d_o, so if the image is on the opposite side (d_i positive) and object is on the front (d_o positive), then m is negative, so inverted. 

Therefore, all the conclusions make sense. 

Let me just check with another method. Suppose I use the lens formula with the numbers again.

1/f = 1/d_o + 1/d_i,

1/10 = 1/15 + 1/d_i,

1/d_i = 1/10 - 1/15,

Convert to decimal to see:

1/10 = 0.1, 1/15 ≈ 0.066666...

0.1 - 0.066666... ≈ 0.033333... which is 1/30, so d_i is